Question

28.69

A 100 μF defibrillator capacitor is charged to 1500 V. When fired through a patient's chest, it loses 95% of its charge in 40 ms.

What is the resistance of the patient's chest?

PS: 133 ohms, 4.98*10^3 are WRONG ANSWERS)

Answer 1

Given

Capacitance        C = 100 x 10^-6 F

Initial voltage     V_o = 1500 V

Time taken         t = 40 x 10^-3 S =

Total discharge is 95%

Solution

The initial charge Q_o = CV₀

Q_o = 100 x 10^-6 x 1500

Q_o = 0.15 C

Charge at any time t during discharge

Q = Q_o(1 – e^-t/RC)

Q_o - 0.95Q_o = Q_o(1-e^-0.04/RC)

0.05Q_o= Q_o(1-e^-0.04/RC)

0.05 = 1-e^-0.04/RC

e^-0.04/RC = 0.95

-0.04/Rc = ln0.95

-0.04/RC = -0.05129

RC = 0.7799

R = 0.7799/100 x 10^-6

R = 7799 ohm